GRE Physics Companion: Damped Harmonic Motion
The standard linear damped oscillator is
with
The characteristic roots are
Figure 1. A compact damping strategy: compute ω0 and γ, classify the roots, choose the correct
solution form, then use energy or peak decay as an independent check.
1 High-value GRE facts
- γ = b∕(2m) for the convention ẍ + 2γẋ + ω02x = 0.
- ω0 =
is the undamped natural angular frequency.
- Underdamped: γ < ω0.
- Critical: γ = ω0.
- Overdamped: γ > ω0.
- ωd =
in the underdamped regime.
- The underdamped envelope decays as e−γt.
- critical damping requires bc = 2
.
- mechanical energy satisfies dE∕dt = −bv2.
- logarithmic decrement is Λ = 2πγ∕ωd.
Part I: Original GRE-style problems
Problem 1: damping rate
For
the damping rate in
is
- b∕m
- b∕(2m)
- 2b∕m
- k∕b

Problem 2: regime classification
A system has
It is
- undamped
- underdamped
- critically damped
- overdamped
- unstable
Problem 3: damped frequency
For the system in Problem 2,
is
- 2 rad/s
- 3 rad/s
- 4 rad/s
- 5 rad/s
rad/s
Problem 4: amplitude decay
For an underdamped oscillator with
the amplitude envelope after
is multiplied by
- e−2
- e−1
- e−1∕2
- 2
- 1
Problem 5: critical coefficient
For
critical damping requires
- 5 kg/s
- 10 kg/s
- 20 kg/s
- 25 kg/s
- 100 kg/s
Problem 6: critical solution
The general critically damped displacement has the form
- A cos ω0t
- Ae−γt cos ω
dt
- (C + Dt)e−ω0t
- Ceω0t + De−ω0t
- C + Dt
Problem 7: energy loss rate
For viscous damping,
equals
- +bv2
- −bv2
- −kx2
- −mgv
- zero
Problem 8: phase plane
The underdamped phase-plane trajectory is approximately
- a closed ellipse
- an outward spiral
- an inward spiral
- a straight line
- a hyperbola
Problem 9: logarithmic decrement
Successive same-sign peaks are
The logarithmic decrement is
- 1∕2
- ln 2
- 2
- e2
- π
Problem 10: overdamped roots
For an overdamped oscillator, the characteristic roots are
- purely imaginary
- complex conjugates with negative real part
- two distinct negative real numbers
- one positive and one negative number
- both zero
Problem 11: weak-damping quality factor
For weak viscous damping,
is approximately
- 2γ∕ω0
- ω0∕(2γ)
- γω0
- ω02∕γ
- 1
Problem 12: damping and period
For an underdamped oscillator,
- ωd > ω0
- ωd = ω0 for all damping
- ωd < ω0
- ωd = 0 always
- ωd is imaginary
Part II: Complete worked solutions
Solution 1
Dividing by mass gives
Therefore
so
Answer: (B).
Solution 2
Because
The system is underdamped. Answer: (B).
Solution 3
| ωd | =  | (19)
|
| =  | (20)
|
| = 4 rad∕s. | (21) |
Answer: (C).
Solution 4
The envelope factor is
Answer: (B).
Solution 5
| bc | = 2 | (23)
|
| = 2 | (24)
|
| = 20 kg∕s. | (25) |
Answer: (C).
Solution 6
A repeated root produces
Answer: (C).
Solution 7
The damping force power is
Thus
Answer: (B).
Solution 8
Damping removes energy, so the closed undamped phase-plane orbit contracts toward the origin. In
the underdamped case it spirals inward. Answer: (C).
Solution 9
Answer: (B).
Solution 10
For
is real and smaller than
Thus both roots are real, distinct, and negative. Answer: (C).
Solution 11
For weak damping,
Answer: (B).
Solution 12
Answer: (C).
2 GRE checklist
- Compute ω0 and γ before selecting a solution form.
- Compare γ directly with ω0.
- Keep the convention 2γ = b∕m consistent.
- For underdamping, use ωd =
.
- For critical damping, remember the extra factor of t in the second independent solution.
- For overdamping, expect two decaying exponentials, not a sinusoid.
- Use dE∕dt = −bv2 as a sign and physics check.
- Use logarithmic decrement only with corresponding same-sign peaks.
References
References
[1] PhysicsLibrary, M06-01, Simple Harmonic Motion.
[2] A. P. French, Vibrations and Waves, W. W. Norton, 1971.
[3] J. R. Taylor, Classical Mechanics, University Science Books, 2005.